Question

Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH. First what is the initial pH (before any NaOH is added)? The Ka for HOCI is 3.0 x 10-8 M. Need help? Hint, given in general feedback. Answer: 3.96 Make ICE table and use simplified method of successive approximations (no iterations needed). How many mL of NaOH are added to reach the equivalence point? Answer: 8 What is the pH after 3.20 mL of NaOH are added? Hint given in general feedback. Answer: 7.6 X

Answer 1

Given that Molarity of HCIO = 0.400 M Ka of HClO = 3.0 x 10-8 pKa = -log(3.0 x 10-8) = 7.522879 = 7.523 Concentration of NaOH = 0.500 M Number of moles of HCIO= M*V = 0.400 M * 10.0 mL = 4.0 mmol At equivalence point, number of moles of HClO= number of moles of NaOH Number of moles of NaOH at equivalence point = 4.0 mmol Volume of NaOH required to reach equivalence point = 4.0 mmol/0.500 M = 8.00 mL

Addition of 3.20 mL of NaOH: Number of moles of HCIO = M*V = 0.400 M * 10.0 mL = 4.0 mmol Number of moles of NaOH = M*V = 0.500 M* 3.20 mL = 1.6 mmol Net moles of HCIO = 4.0 - 1.6 = 2.4 mmol Kb mmol HCIO + NaOH = Nacio + H20 Initial change Equilibrium 4.0 -1.6 0 -1.6 0 +1.6 2.4 1.6 [base] _7523+ log According to Henderson's equation 1.6 pH = pK, + log = 7.523+ log = (acid] 2.4 pH = 7.35 = 7.523 -0.17609

At equivalence point: Volume of NaOH to reach equivalence point = 8.00 mL At this point, there is no HClO is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate base. moles 4.00 mmol [CIO-]= -= 0.2222M Total volume (10.0+8.0)mL 1x10-14 K, for ClO = 3.0x10-8 -3.33 x 10-7 Kb M CIO + H2O = HCIO + OH- Initial 0.2222 Change Equillibrium 0.2222-X

X.X -= 3.33x10-7 0.2222-X x is small, so 0.2222-x=0.2222 3.X.X_-33310- 0.2222 = 3.33x10- x² = 7.41x10-8 x=0.000272M :: [OH"]=0.000272M pOH = -log(0.000272)= 3.57 pH =14 – POH =14 – 3.57 = 10.43 Addition of 8.40 mL of NaOH: Number of moles of HCIO = M*V = 0.400 M * 10.0 mL = 4.0 mmol Number of moles of NaOH= M*V = 0.500 M * 8.40 mL = 4.2 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted NaOH in the solution. Net moles of NaOH = 4.2-4.0 = 0.2 mmol Total volume of the solution = 10.0 + 8.40 = 18.4 mL Conc. of NaOH = [H+] = moles/ volume = 0.2 mmol / 18.4 mL = 0.01087 M pOH =-log [OH-] = -log[0.01087] = 1.96 pH = 14 - pOH = 14 – 1.96 = 12.04